If the lift starts moving up with a uniform speed of $5 m$ $s^{- 1}$ and the girl again throws the ball up with the same speed, how long does the ball take to return to her hands?

.5 s

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1.0 s

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1.5 s

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2.0 s

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Solution

The correct option is B $1.0 s$
Given,
Velocituy of Ball relative to lift:
$u_{r} = 5 m / s$
$a_{r} = g = - 10 m / s^{2}$
From the 1st equation of motion,
$v = u_{r} + a_{r} t$
$0 = 5 - 10 t$
$t = \frac{5}{10} = 0.5 s e c$
Time of flight, $T = 2 t$
$T = 2 \times 0.5$
$T = 1 s e c$

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A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

Q. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s¯¹. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s¯¹ and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?

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Q. A girl standing on a pedestal at rest throws a ball upwards with maximum possible speed of

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. If the platform starts moving up at

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and the girl again throws the similar ball in similar way. The time taken in the previous case and the time taken in the second case to return to her hands are

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If the lift starts moving up with a uniform speed of 5m s−1 and the girl again throws the ball up with the same speed, how long does the ball take to return to her hands?

The correct option is B 1.0sGiven,Velocituy of Ball relative to lift:ur=5m/sar=g=−10m/s2From the 1st equation of motion, v=ur+art0=5−10tt=510=0.5secTime of flight, T=2tT=2×0.5T=1sec

If the lift starts moving up with a uniform speed of $5 m$ $s^{- 1}$ and the girl again throws the ball up with the same speed, how long does the ball take to return to her hands?

The correct option is B $1.0 s$
Given,
Velocituy of Ball relative to lift:
$u_{r} = 5 m / s$
$a_{r} = g = - 10 m / s^{2}$
From the 1st equation of motion,
$v = u_{r} + a_{r} t$
$0 = 5 - 10 t$
$t = \frac{5}{10} = 0.5 s e c$
Time of flight, $T = 2 t$
$T = 2 \times 0.5$
$T = 1 s e c$